What is the theoretical yield of sodium chloride for the reaction of 36.0 g Na with 75.2 g Cl2 of the following 2NA+Cl2---> 2NaCl? I keep getting stuck with my conversions. I started with 36.0g NA (1 mol Na/22.99g Na) (1 mol Cl2/2mol Na) =0.782 mol Cl2.

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What is the theoretical yield of sodium chloride for the reaction of 36.0 g Na with 75.2 g Cl2 of the following 2NA+Cl2---> 2NaCl? I keep getting stuck with my conversions. I started with 36.0g NA (1 mol Na/22.99g Na) (1 mol Cl2/2mol Na) =0.782 mol Cl2.

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Answer 1 · 2014-09-29T00:14:43

The theoretical yield is 91.5g of $N a C l$ $NaCl$

Method 1

Start with the equation:

$2 N a_{(s)} + C l_{2 (g)} \to 2 N a C l_{(s)}$ $2Na_((s))+Cl_(2(g))rarr2NaCl_((s))$

So 2 mole $N a$ $Na$ reacts with 1 mole $C l_{2}$ $Cl_2$ to give 2 moles $N a C l$ $NaCl$

Convert moles to grams: $A_{r} N a = 23$ $A_r Na=23$ $A_{r} C l = 35.5$ $A_rCl=35.5$

2x23 = 46g $N a$ $Na$ reacts with 2x 35.5 = 71g $C l_{2}$ $Cl_2$ to give 2x(23 +35.5 ) = 117g $N a C l$ $NaCl$

Use simple proportion to find how much would be made from 36g $N a$ $Na$ by first working out what 1 g of $N a$ $Na$ would give by dividing through by 46:

(46/46)g $N a$ $Na$ reacts with (71/46) g $C l_{2}$ $Cl_2$ to give (117/46)g $N a C l$ $NaCl$

So:

1g $N a$ $Na$ reacts with 1.54g $C l_{2}$ $Cl_2$ to give 2.54g $N a C l$ $NaCl$

So:

36g $N a$ $Na$ reacts with 1.54x36= 55.44g $C l_{2}$ $Cl_2$ to give 2.54x36=91.44g $N a C l$ $NaCl$

You can see that the chlorine is in excess so the theoretical yield is 91.5g of $N a C l$ $NaCl$ to 1sf.

Method 2

Start with the balanced equation as above. Calculate the amount of NaCl that can form from each reactant.

36.0 g Na × $\frac{1 mol Na}{22.99 g Na} \times \frac{2 mol NaCl}{2 mol Na}$ $"1 mol Na"/"22.99 g Na" × "2 mol NaCl"/"2 mol Na"$ = 1.566 mol NaCl (3 significant figures + 1 guard digit)

75.2 g Cl₂ × $\frac{{1 mol Cl}_{2}}{{70.91 g Cl}_{2}} \times \frac{2 mol NaCl}{{1 mol Cl}_{2}}$ $("1 mol Cl"_2)/("70.91 g Cl"_2) × "2 mol NaCl"/("1 mol Cl"_2)$ = 2.121 mol NaCl

The Na gives fewer moles of NaCl, so Na is the limiting reactant.

The theoretical yield is

1.566 mol NaCl × $\frac{58.44 g NaCl}{1 mol NaCl} = 91.5 g NaCl$ $"58.44 g NaCl"/"1 mol NaCl" =" 91.5 g NaCl"$

Method 3

Divide the moles of each reactant by its coefficient in the balanced equation.

36.0 g Na × $\frac{1 mol Na}{22.99 g Na} = 1.566 mol Na$ $"1 mol Na"/"22.99 g Na" =" 1.566 mol Na"$ (3 significant figures + 1 guard digit)

75.2 g Cl₂ × $\frac{{1 mol Cl}_{2}}{{70.91 g Cl}_{2}} = {2.121 mol Cl}_{2}$ $("1 mol Cl"_2)/("70.91 g Cl"_2) =" 2.121 mol Cl"_2$

$2 Na + {Cl}_{2} \to 2 NaCl$ $2"Na" + "Cl"_2 → 2"NaCl"$

$\frac{1.566}{2}$ $1.566/2$ ; $\frac{2.121}{1}$ $2.121/1$

0.7829; 2.121

NaCl gives fewer "moles of reaction", so NaCl is the limiting reactant.

So theoretical yield is

0.7829 mol reaction × $\frac{2 mol NaCl}{1 mol reaction} \times \frac{58.44 g NaCl}{1 mol NaCl}$ $"2 mol NaCl"/"1 mol reaction" × "58.44 g NaCl"/"1 mol NaCl"$ = 91.5 g NaCl

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What is the theoretical yield of sodium chloride for the reaction of 36.0 g Na with 75.2 g Cl2 of the following 2NA+Cl2---> 2NaCl? I keep getting stuck with my conversions. I started with 36.0g NA (1 mol Na/22.99g Na) (1 mol Cl2/2mol Na) =0.782 mol Cl2.

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